Step-by-step explanation:
if x + 1 and x - 1 are factors, then (x + 1)(x - 1) = (x² - 1) is a factor. and vice versa.
these terms are factors, if they can divide the main expression without any remainder.
x⁴ + ax³ + 2x² - 3x + b / x² - 1 = x² + ax + 3
- (x⁴ - x²)
------------------------
0 + ax³ + 3x² - 3x + b
- (ax³ - ax)
-----------------------------------
0 + 3x² - (3 - a)x + b
- (3x² - 3)
------------------------------------------
0 - (3 - a)x + (b + 3) = 0
so, a = 3, b = -3
test :
x⁴ + 3x³ + 2x² - 3x - 3 / x² - 1 = x² + 3x + 3
- (x⁴ - x²)
------------------------
0 + 3x³ + 3x² - 3x - 3
- (3x³ - 3x)
-----------------------------------
0 + 3x² - (3 - 3)x - 3
- (3x² - 3)
------------------------------------------
0 - 0 - 0 = 0
(x² + 3x + 3)(x + 1) = x³ + x² + 3x² + 3x + 3x + 3 =
= x³ + 4x² + 6x + 3
(x³ + 4x² + 6x + 3)(x - 1) = x⁴ - x³ + 4x³ - 4x² + 6x² - 6x + 3x - 3 =
= x⁴ + 3x³ + 2x² - 3x - 3
so, yes, the division by (x² - 1) has 0 remainder, and when multiplying that result by the factors (x + 1) and (x - 1) we get the full original expression.
